-26=-15t+t^2

Solution for -26=-15t+t^2 equation:

-26=-15t+t^2

We move all terms to the left:

-26-(-15t+t^2)=0

We get rid of parentheses

-t^2+15t-26=0

We add all the numbers together, and all the variables

-1t^2+15t-26=0

a = -1; b = 15; c = -26;
Δ = b2-4ac
Δ = 152-4·(-1)·(-26)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-11}{2*-1}=\frac{-26}{-2}
=+13
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+11}{2*-1}=\frac{-4}{-2}
=+2
$